记b=1/a,原已知式变为
\(x+1+\frac{1}{x}=b\)。即\(x+\frac{1}{x}=b-1\)。两边平方,并把1移到右端得
\(x^2+1+\frac{1}{x^2}=(b-1)^2-1=b^2-2b\).
所求式子上下除以\(x^2\),变为
\(\frac{1}{x^2+1+1/x^2}=\frac{1}{b^2-2b}=\frac{a^2}{1-2a}\)
\(x+1+\frac{1}{x}=b\)。即\(x+\frac{1}{x}=b-1\)。两边平方,并把1移到右端得
\(x^2+1+\frac{1}{x^2}=(b-1)^2-1=b^2-2b\).
所求式子上下除以\(x^2\),变为
\(\frac{1}{x^2+1+1/x^2}=\frac{1}{b^2-2b}=\frac{a^2}{1-2a}\)
锟斤拷锟洁辑时锟斤拷: 2021-10-12 00:25:34